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lisabon 2012 [21]
3 years ago
7

Velocity and acceleration are both vectors; they have a direction. What is thedirection of the velocity and acceleration vectors

of a tennis ball that is rising upward towards its highest point above the ground? a.The velocity vector is directed upward. b.The velocity vector is directed downward. c.The acceleration is directed downward. d.Both A & C
Physics
1 answer:
just olya [345]3 years ago
3 0

Answer:

d. Both A & C

Explanation:

- The velocity of the ball is a vector, whose magnitude indicates its rate of change of position, and it has a direction. In this case, the ball is moving upward, therefore the direction of motion is upward, so the direction of the velocity is upward as well.

- The acceleration of the ball is a vector, whose magnitude indicates the rate of change of the velocity. The direction of the acceleration is:

-- positive if the the magnitude of the velocity is increasing

-- negative if the magnitude of the velocity is decreasing

For a ball thrown upward, the acceleration is given by the acceleration of gravity, g=9.8 m/s^2. This acceleration points downward, and it is constant during the entire motion. In particular, it does not change direction, as it is always directed downward. Therefore, the acceleration of the ball is downward.

So, the correct answer is

d.Both A & C

Since A and C are both true:

a.The velocity vector is directed upward.

c.The acceleration is directed downward.

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Answer:

5.444\times 10^{-4}

Explanation:

The momentum of the neutron before and after the decay  is the same since there's no external force.

P_{sys}=const\\\\P=mv\\\\K=0.5mv^2

#The neutron is initially at rest, so after the decay:

P_A+P_B=0\\\\P_A=-P_B

#After decay, the proton has +ve direction  with a velocity v_Awhile the electron moves in a negative direction with a velocity v_B

Therefore:

P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B

Let the energy released during the decay be Q:

Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}

Hence,Kp/Ktot is 5.444x10^(-4)

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Explanation:

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